Natural Variables of the Thermodynamic Potentials

Imagine you know just one function of a system. From its shape alone, you can predict its pressure, entropy, heat capacity, and squeeze-response. That is the promise of a thermodynamic potential expressed in its natural variables. Pick the wrong variables, though, and the same function becomes awkward and less informative. This idea matters because chemists rarely control energy directly. We control temperature and pressure in a beaker. Or we control temperature and volume in a bomb calorimeter. Matching the potential to the variables we actually manipulate matters greatly. It turns thermodynamics from bookkeeping into prediction. This article builds the four potentials — $U$ , $H$ , $A$ , $G$ — from the first law. You will see why each one prefers a specific pair of variables. Then you will see how a single differentiation reproduces entropy, volume, pressure, and more. The payoff is the toolkit behind Maxwell relations. It also explains the temperature and pressure dependence of $G$ .

Key takeaways

Each thermodynamic potential is a function of two specific "natural" variables, fixed by how it was constructed from the first law.
Differentiating a potential with respect to its natural variables regenerates every other state function — no new physics needed.
$U(S,V)$ , $H(S,p)$ , $A(T,V)$ , $G(T,p)$ are the four standard potentials, related by Legendre transforms.
Using non-natural variables still gives a valid function, but it no longer delivers the full state of the system from one differentiation.
$G(T,p)$ is the chemist's workhorse because $T$ and $p$ are the variables actually controlled at the bench.
Key idea: A thermodynamic potential is "complete" only when written in its natural variables — that pairing is what lets one function generate an entire property set by differentiation alone.

The core insight

The four thermodynamic potentials are not arbitrary combinations of $U$ , $S$ , $T$ , $p$ , $V$ ; each is built by a Legendre transform designed so that its total differential naturally involves one particular pair of variables, and only when written in that pair does differentiating it once yield every remaining thermodynamic property without additional integration.

The physical picture

Start from the fundamental equation of the first and second laws for a closed system doing only $pV$ work:

dU = TdS - pdV

This equation says something precise: if you know $U$ as a function of $S$ and $V$ , then $T$ and $p$ fall out immediately as slopes of $U(S,V)$ . That is not true if you know $U$ as a function of, say, $T$ and $V$ — you would need extra experimental input (a heat capacity, an equation of state) to get $T$ and $p$ back. $S$ and $V$ are $U$ 's natural variables. But chemists rarely hold $S$ and $V$ fixed. We hold $T$ and $p$ fixed at the bench, or $T$ and $V$ fixed in a rigid, thermostatted vessel. So we perform Legendre transforms on $U$ — swapping an extensive variable for its conjugate intensive one — to build three new potentials whose natural variables match what we actually control:

Enthalpy $H = U + pV$ , natural variables $(S,p)$
Helmholtz energy $A = U - TS$ , natural variables $(T,V)$
Gibbs energy $G = U + pV - TS = H - TS$ , natural variables $(T,p)$

Each transform trades one variable in the differential for its conjugate partner, without losing any information — the transform is exact and reversible, like changing coordinates on a map without deleting any streets.

Figure

The Legendre-transform square: each arrow trades one variable for its conjugate

Once a potential is written in its natural variables, its total differential has a completely fixed form, and the coefficients of $dS$ , $dV$ , $dT$ , $dp$ are exact partial derivatives that hand you back $T$ , $p$ , $-S$ , $V$ directly — no extra measurement needed. That single fact is the seed of the Maxwell relations covered next.

Figure

Slopes of G(T,p) read off directly as −S and V

The core relations

The fundamental equation and the three Legendre-transformed potentials, each with its natural-variable differential:

dU = TdS - pdV

where: $U(S,V)$ ; slopes give $T = (\partial U/\partial S)_V$ and $-p = (\partial U/\partial V)_S$ .

dH = TdS + Vdp

where: $H = U+pV$ , natural variables $(S,p)$ ; $T=(\partial H/\partial S)_p$ , $V=(\partial H/\partial p)_S$ .

dA = -SdT - pdV

where: $A = U-TS$ , natural variables $(T,V)$ ; $-S=(\partial A/\partial T)_V$ , $-p=(\partial A/\partial V)_T$ .

dG = -SdT + Vdp

where: $G = H-TS$ , natural variables $(T,p)$ ; $-S=(\partial G/\partial T)_p$ , $V=(\partial G/\partial p)_T$ .

\left(\frac{\partial G}{\partial T}\right)_p = -S, \qquad \left(\frac{\partial G}{\partial p}\right)_T = V

where: these two derivatives are the basis for the temperature and pressure dependence of $G$ discussed next in this chapter.

Derivation ABuilding H, A, G from U by Legendre transform

Goal. Show that enthalpy, Helmholtz energy, and Gibbs energy are forced into their natural variables by construction, not by convention.

Step 1. Begin from the combined first and second law for a closed, $pV$ -work-only system:

dU = TdS - pdV

This already tells us $U$ is naturally a function of $S$ and $V$ : its differential contains only $dS$ and $dV$ .

Step 2. Suppose we want a potential whose natural variable is $p$ instead of $V$ , because pressure — not volume — is what we control in an open beaker. Define

H \equiv U + pV

Differentiate using the product rule:

dH = dU + pdV + Vdp

Substitute $dU = TdS - pdV$ :

dH = TdS - pdV + pdV + Vdp = TdS + Vdp

The $pdV$ terms cancel exactly. $H$ 's differential now contains only $dS$ and $dp$ : $H(S,p)$ is complete in those variables.

Step 3. Now trade $S$ for $T$ , since temperature is far easier to control than entropy. Define

A \equiv U - TS

Differentiate:

dA = dU - TdS - SdT = (TdS - pdV) - TdS - SdT = -SdT - pdV

Again the matching terms cancel, leaving $dA$ in terms of $dT$ and $dV$ only: $A(T,V)$ is complete.

Step 4. Perform both trades at once. Define

G \equiv U + pV - TS = H - TS = A + pV

Differentiate any equivalent form, e.g. from $H - TS$ :

dG = dH - TdS - SdT = (TdS+Vdp) - TdS - SdT = -SdT + Vdp

Every cross term cancels, leaving exactly two terms: $dG(T,p) = -SdT + Vdp$ .

Step 5. Read off the state functions as partial derivatives, since each differential is now exact in its two natural variables:

\left(\frac{\partial U}{\partial S}\right)_V = T,\quad \left(\frac{\partial U}{\partial V}\right)_S=-p

\left(\frac{\partial H}{\partial S}\right)_p = T,\quad \left(\frac{\partial H}{\partial p}\right)_S=V

\left(\frac{\partial A}{\partial T}\right)_V = -S,\quad \left(\frac{\partial A}{\partial V}\right)_T=-p

\left(\frac{\partial G}{\partial T}\right)_p = -S,\quad \left(\frac{\partial G}{\partial p}\right)_T=V

Boxed result.

\boxed{U(S,V),\ H(S,p),\ A(T,V),\ G(T,p) \text{ each yield } T, p, S, V \text{ by a single differentiation in their natural variables.}}

Worked examples

Example 1 — reading properties off G(T,p) for an ideal gas. For 1.00 mol of an ideal gas, the molar Gibbs energy relative to a reference state $(T^\circ, p^\circ)$ can be written using $(\partial G/\partial p)_T = V = RT/p$ :

G_m(T,p) = G_m^\circ + RT\ln\frac{p}{p^\circ}

Take $T = 298.15\ \text{K}$ , $p^\circ = 1.00\ \text{bar}$ , and evaluate at $p = 5.00\ \text{bar}$ . Step 1 — compute $RT$ :

RT = (8.314\ \text{J K}^{-1}\text{mol}^{-1})(298.15\ \text{K}) = 2478.8\ \text{J mol}^{-1}

Step 2 — compute the log term:

\ln\frac{5.00}{1.00} = \ln(5.00) = 1.6094

Step 3 — combine:

G_m(298.15\ \text{K}, 5.00\ \text{bar}) - G_m^\circ = (2478.8\ \text{J mol}^{-1})(1.6094) = 3989\ \text{J mol}^{-1} = 3.99\ \text{kJ mol}^{-1}

Step 4 — now use the other natural-variable derivative to get the entropy change on the same path at fixed $T$ . Since $(\partial G/\partial p)_T = V$ was already used above, instead check consistency using $(\partial G/\partial T)_p = -S$ : because $G_m^\circ$ carries all the $T$ -dependence not from the $\ln p$ term, differentiating the full expression with respect to $T$ at fixed $p$ gives

-S_m(T,p) = \left(\frac{\partial G_m^\circ}{\partial T}\right) + R\ln\frac{p}{p^\circ}

so raising pressure at constant temperature lowers the molar entropy by $R\ln(p/p^\circ) = (8.314)(1.6094) = 13.4\ \text{J K}^{-1}\text{mol}^{-1}$ , exactly consistent with compressing an ideal gas. Sanity check: increasing $p$ raises $G$ (more confined, less probable states) and lowers $S$ (fewer accessible microstates) — both signs match physical intuition, and both came from differentiating one function, $G(T,p)$ , in its two natural variables.

Example 2 — extracting $C_p$ and the isothermal compressibility from G(T,p), applied to liquid water. The full power of natural variables shows up in second derivatives. Since $S = -(\partial G/\partial T)_p$ , the constant-pressure heat capacity follows from a second derivative of $G$ :

C_p = T\left(\frac{\partial S}{\partial T}\right)_p = -T\left(\frac{\partial^2 G}{\partial T^2}\right)_p

and since $V = (\partial G/\partial p)_T$ , the isothermal compressibility follows similarly:

\kappa_T = -\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_T = -\frac{1}{V}\left(\frac{\partial^2 G}{\partial p^2}\right)_T

Take liquid water at $298.15\ \text{K}$ and $1.00\ \text{bar}$ , molar volume $V_m = 18.07\ \text{cm}^3\text{mol}^{-1} = 1.807\times10^{-5}\ \text{m}^3\text{mol}^{-1}$ , with a measured isothermal compressibility $\kappa_T = 4.53\times10^{-5}\ \text{bar}^{-1}$ . Step 1 — convert $\kappa_T$ to SI (Pa $^{-1}$ ), since $1\ \text{bar} = 10^5\ \text{Pa}$ :

\kappa_T = 4.53\times10^{-5}\ \text{bar}^{-1} \times \frac{1\ \text{bar}}{10^5\ \text{Pa}} = 4.53\times10^{-10}\ \text{Pa}^{-1}

Step 2 — this means the second $p$ -derivative of $G_m$ at fixed $T$ is

\left(\frac{\partial^2 G_m}{\partial p^2}\right)_T = -V_m\kappa_T = -(1.807\times10^{-5}\ \text{m}^3\text{mol}^{-1})(4.53\times10^{-10}\ \text{Pa}^{-1}) = -8.19\times10^{-15}\ \text{m}^3\ \text{Pa}^{-1}\text{mol}^{-1}

a tiny number, telling us water's volume is nearly insensitive to pressure near 1 bar — exactly why liquids are called "incompressible" to a good approximation. Step 3 — for the heat capacity direction, use the known $C_{p,m} = 75.3\ \text{J K}^{-1}\text{mol}^{-1}$ for liquid water at 298 K to back out the curvature of $G_m(T)$ at fixed $p$ :

\left(\frac{\partial^2 G_m}{\partial T^2}\right)_p = -\frac{C_{p,m}}{T} = -\frac{75.3\ \text{J K}^{-1}\text{mol}^{-1}}{298.15\ \text{K}} = -0.2525\ \text{J K}^{-2}\text{mol}^{-1}

Sanity check: $C_p>0$ requires this curvature to be negative, i.e. $G_m(T)$ is concave down at fixed $p$ — consistent with entropy $S=-(\partial G/\partial T)_p$ increasing with $T$ , as it must for any stable substance.

Limiting cases & common traps

Situation	What natural variables tell you	Common error
Constant $S,V$ process	$U$ decreases toward equilibrium	Trying to minimize $U$ at constant $T,p$ instead — wrong potential
Constant $S,p$ process	$H$ decreases toward equilibrium	Confusing $H$ 's natural variable $S$ with the everyday constant- $T$ lab condition
Constant $T,V$ process (e.g. bomb calorimeter)	$A$ decreases toward equilibrium; $A$ = maximum non-expansion + expansion work available	Using $G$ for a rigid, sealed vessel where volume, not pressure, is fixed
Constant $T,p$ process (beaker chemistry)	$G$ decreases toward equilibrium; $\Delta G$ predicts spontaneity	Forgetting $G$ demands both $T$ and $p$ constant, not just one
Writing $U(T,V)$ or $G(S,V)$	Still mathematically a function, but a single differentiation no longer regenerates every property	Assuming any parametrization of a potential is "as good as" the natural one

The chemist behind it

Josiah Willard Gibbs, working largely alone at Yale in the 1870s, wrote "On the Equilibrium of Heterogeneous Substances" — a paper so dense and far ahead of contemporary chemistry that it took decades for European scientists like Ostwald and van der Waals to popularize it. Gibbs recognized that different experimental constraints demand different potentials, and he systematically worked out the Legendre-transform relationships between them, essentially inventing the entire multivariable framework this article describes. The Gibbs energy — the potential natural to constant temperature and pressure, exactly the conditions of ordinary bench chemistry — is his namesake precisely because he identified why it, and not $U$ alone, governs spontaneity and equilibrium in the laboratory.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American mathematical physicist who founded chemical thermodynamics and statistical mechanics, largely from New Haven, Connecticut

Ask the tutor: How do these same natural-variable derivatives combine to generate the four Maxwell relations?

Why does each potential have exactly two natural variables, not one or three

For a closed system doing only $pV$ work, the state is fully fixed by two independent variables. The first law differential $dU = TdS - pdV$ has exactly two terms, so $U$ is naturally a function of two variables, $S$ and $V$ . Every Legendre transform preserves that count: it swaps one variable for its conjugate but never adds or removes a degree of freedom. Hence $H$ , $A$ , and $G$ each keep exactly two natural variables too.

Can I still calculate G at some T and V, not p

Yes, $G$ is a well-defined state function evaluated at any state, however you specify that state. But if you specify $G(T,V)$ , differentiating it with respect to $V$ at fixed $T$ does not simply hand you $-p$ or any other clean thermodynamic quantity, because $V$ is not conjugate to a term appearing in $G$ 's natural differential $dG = -SdT + Vdp$ . You would need extra information, like an equation of state, to convert between representations.

How does this connect to Maxwell relations

Because each potential's differential is exact in its natural variables, mixed second partial derivatives must be equal regardless of the order of differentiation. Applying that equality to $dU$ , $dH$ , $dA$ , and $dG$ in turn generates the four Maxwell relations, each connecting a hard-to-measure entropy derivative to an easy-to-measure volume, pressure, or temperature derivative.

Why is A called the "work function"

At constant temperature, the decrease in Helmholtz energy equals the maximum total work — expansion plus non-expansion — that a process can extract, which is why the German literature calls it "Arbeit" (work). Its natural variables $(T,V)$ make it the natural choice for constant-volume, constant-temperature setups like a bomb calorimeter or a rigid sealed reactor.

Does the choice of natural variables affect the numerical value of a potential

No. $G$ , $A$ , $H$ , $U$ at a given physical state have fixed values — a state function's value never depends on how you chose to parametrize it mathematically. What changes is how easily you can extract other properties by differentiation. Natural variables make that extraction a one-step, exact process; other variable choices require extra experimental input.